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I am studying how to get the normalization factor algebraically in the exited states of the harmonic oscillator using the beautiful ladder operators technique.

I am stuck at the point where is stated that the product of the ladder operators yields the energy levels (in $\widehat{a}^\dagger\widehat{a}$ case) and the energy levels + 1 (in $\widehat{a}\widehat{a}^\dagger$ case).

$$\widehat{a}\widehat{a}^\dagger\psi_n= (n+1)\psi_n$$

$$\widehat{a}^\dagger\widehat{a}\psi_n=n\psi_n$$

The thing is that I know how to get them mathematically using:

$$\widehat{a}^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$$

$$\widehat{a}|n\rangle=\sqrt{n}|n-1\rangle$$

But I am not satisfied with this due to the fact that my textbook (Griffiths) does not obtain them like that but using both the Schrodinger Equation and the solution to the exited states of the harmonic oscillator:

$$\hbar\omega\left(a\pm a\mp \pm \frac{1}{2}\right)\psi = E \psi$$

$$\psi_n = A_n (\widehat{a}^\dagger )^n \psi_0$$

But I am not getting them using the stated way. May you please prove it?

I am also wondering why the product of the two ladder operators do yield energy levels without a constant of proportionality. In fact; when just one operator is multiplied by the eigenstates you get a constant of proportionality multiplied by the eigenstates (+1 or -1; depending on the operator we used). But this may be another question so please let me know whether I should post it separately:

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closed as off-topic by Chris, Cosmas Zachos, user191954, Kyle Kanos, glS Nov 10 '18 at 13:46

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  • $\begingroup$ @Qmechanic From my point of view, this is not a homework-and-exercise question. I was reading the book and got stuck in one of the steps. Does that count as a homework like question on PSE? $\endgroup$ – JD_PM Nov 9 '18 at 18:11
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    $\begingroup$ Hi JD_PM. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Nov 9 '18 at 18:14
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I think you misunderstood what he meant. Griffiths did use one of the equations you stated : \begin{equation} \hbar\omega\left(a\pm a\mp \pm \frac{1}{2}\right)\psi = E \psi \ , \end{equation} but the other equation he used (2.61 in my edition) wasn't the form of the eigenstates, but the eigenvalues which are written beside it, and not numbered separetedly: \begin{equation} \hbar\omega\left(n +\frac{1}{2}\right) = E_n. \end{equation} Substitution of this equation, as well as the Hamiltonian $H=\hbar\omega\left(a_+a_- + \frac{1}{2}\right)$ on the first gives the result immediately. a_+a_- is found by direct inspection and the other equation can be infered, for example, from the commutation relation $[a_+,a_-]=1$

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